我的目標(biāo)是在每次迭代中將多個(gè)熊貓數(shù)據(jù)幀連接成單個(gè)數(shù)據(jù)幀。我正在抓取一個(gè)表并用它創(chuàng)建數(shù)據(jù)幀。下面是注釋的代碼。def visit_table_links():    links = grab_initial_links()    df_final = None    for obi in links:        resp = requests.get(obi[1])        tree = html.fromstring(resp.content)        dflist = []        for attr in tree.xpath('//th[contains(normalize-space(text()),  "sometext")]/ancestor::table/tbody/tr'):            population = attr.xpath('normalize-space(string(.//td[2]))')            try:                population = population.replace(',', '')                population = int(population)                year = attr.xpath('normalize-space(string(.//td[1]))')                year = re.findall(r'\d+', year)                year = ''.join(year)                year = int(year)                #appending a to a list, 3 values first two integer last is string                dflist.append([year, population, obi[0]])            except Exception as e:                pass        #creating a dataframe which works fine        df = pd.DataFrame(dflist, columns = ['Year', 'Population', 'Municipality'])        #first time df_final is none so just make first df = df_final        #next time df_final is previous dataframe so concat with the new one        if df_final != None:            df_final = pd.concat(df_final, df)        else:            df_final = dfvisit_table_links()這是即將到來的數(shù)據(jù)幀第一個(gè)數(shù)據(jù)幀   Year  Population Municipality0  1970       10193   Cape Coral1  1980       32103   Cape Coral2  1990       74991   Cape Coral3  2000      102286   Cape Coral4  2010      154305   Cape Coral5  2018      189343   Cape Coral我已經(jīng)搜索了很多線程并耗盡了我的資源，我是熊貓的新手，不明白為什么會(huì)發(fā)生這種情況，首先，我認(rèn)為這是因?yàn)橹貜?fù)的索引，然后我使用相同的錯(cuò)誤將 uuid.uuid4.int（）作為索引。df.set_index('ID', drop=True, inplace=True)任何指導(dǎo)都將非常有幫助，謝謝。編輯： 1很抱歉沒有明確錯(cuò)誤是從df_final = pd.concat(df_final, df)當(dāng)我嘗試將當(dāng)前數(shù)據(jù)幀與以前的數(shù)據(jù)幀連接時(shí)編輯 2：將參數(shù)作為列表傳遞df_final = pd.concat([df_final, df])仍然相同的錯(cuò)誤